obtaine pdnf of (p->q) ^ (p <->r) without using truth table 

To obtain the **Principal Disjunctive Normal Form (PDNF)** of the expression \((p \to q) \land (p \leftrightarrow r)\) without using a truth table, we need to proceed through logical manipulations. Here's a step-by-step process:

### Step 1: Express the implications and biconditional in terms of basic logical connectives.

- The **implication** \(p \to q\) can be rewritten as:
  \[
  p \to q \equiv \neg p \lor q
  \]
  
- The **biconditional** \(p \leftrightarrow r\) can be rewritten as:
  \[
  p \leftrightarrow r \equiv (p \land r) \lor (\neg p \land \neg r)
  \]

Thus, the original expression becomes:
\[
(\neg p \lor q) \land ((p \land r) \lor (\neg p \land \neg r))
\]

### Step 2: Apply distribution of conjunction over disjunction.

We distribute \((\neg p \lor q)\) over \((p \land r) \lor (\neg p \land \neg r)\), using the distributive law of logic: 
\[
A \land (B \lor C) \equiv (A \land B) \lor (A \land C)
\]

So,
\[
(\neg p \lor q) \land ((p \land r) \lor (\neg p \land \neg r)) \equiv ((\neg p \lor q) \land (p \land r)) \lor ((\neg p \lor q) \land (\neg p \land \neg r))
\]

### Step 3: Simplify each conjunction.

#### First term: \((\neg p \lor q) \land (p \land r)\)
- Distribute \((\neg p \lor q)\) over \((p \land r)\):
  \[
  (\neg p \lor q) \land (p \land r) \equiv (\neg p \land p \land r) \lor (q \land p \land r)
  \]
  
- The term \(\neg p \land p\) is always false, so it simplifies to:
  \[
  (q \land p \land r)
  \]

#### Second term: \((\neg p \lor q) \land (\neg p \land \neg r)\)
- Distribute \((\neg p \lor q)\) over \((\neg p \land \neg r)\):
  \[
  (\neg p \lor q) \land (\neg p \land \neg r) \equiv (\neg p \land \neg p \land \neg r) \lor (q \land \neg p \land \neg r)
  \]
  
- The term \(\neg p \land \neg p \land \neg r\) simplifies to:
  \[
  \neg p \land \neg r
  \]
  
- So the whole expression becomes:
  \[
  (\neg p \land \neg r) \lor (q \land \neg p \land \neg r)
  \]

### Step 4: Combine the results.

Now, we combine the two simplified terms from Step 3:
\[
(q \land p \land r) \lor (\neg p \land \neg r) \lor (q \land \neg p \land \neg r)
\]

### Step 5: Further simplifications (if any).

Observe that \((q \land \neg p \land \neg r)\) is already included in \((\neg p \land \neg r)\) because \(\neg p \land \neg r\) doesn't depend on \(q\). So, the term \(q \land \neg p \land \neg r\) is redundant and can be removed.

Thus, the final PDNF is:
\[
(q \land p \land r) \lor (\neg p \land \neg r)
\]

This is the Principal Disjunctive Normal Form (PDNF) of the expression \((p \to q) \land (p \leftrightarrow r)\).

---

Do you have any questions or would you like more details on any step?

### Related Questions:
1. What is the difference between PDNF and CNF?
2. How do we simplify logical expressions using Karnaugh maps?
3. Can the logical equivalence \((p \leftrightarrow r)\) be simplified differently?
4. How do we handle implications and biconditionals in logical expressions?
5. What is the role of distributive and associative laws in simplifying Boolean expressions?

### Tip:
Always break down logical connectives into simpler forms (like AND, OR, and NOT) when transforming expressions into PDNF or CNF to make distribution and simplification easier.

obtain pdnf of (p->q) ^ (p <->r) without using truth table

Learn how to obtain the Principal Disjunctive Normal Form (PDNF) of the expression (p -> q) ∧ (p <-> r) using logical manipulations, including the use of implication and biconditional transformations. This step-by-step guide covers advanced propositional logic concepts suitable for college-level courses in logic or discrete mathematics.

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